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andreyandreev [35.5K]
3 years ago
13

What happens when you shuffle across a carpeted floor?

Physics
1 answer:
Alex73 [517]3 years ago
3 0
Youhh pick upp electrons from the floor and become positively charged
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We drive at a speed of 20 km/h for 3 hours. Then we drive 4 hours at 30 km/h. Calculate our average speed.
Afina-wow [57]

First speed = 20km/h

Time = 3 hours

Distance = 3×20

<h3> = <u>60 km</u></h3>

Second speed = 30km/h

Time = 4 hours

Distance = 4×30

<h3> = <u>120 km</u></h3>

Total distance = 60+120 = <u>180km</u>

Total time = 3+4 =<u> 7 hours</u>

Average speed = 180/7

<h3> = <u>25.71</u><u> </u><u>km</u><u>/</u><u>h</u></h3>

Hope this will help...

4 0
3 years ago
Read 2 more answers
A constant force is exerted for a short time interval on a cart that is initially at rest on an air track. This force gives the
Ilya [14]

Answer:

d) is the same as when it started from rest

Explanation:

using equation of motion

v = u + at

second law of momentum defines

F = ma

a = F /m

the equation becomes

v = u + (F/m)t

from hear

since v is directly proportional to the force and the force remain the same, the increase in the cart speed will also remain the same.

5 0
3 years ago
Olivia is on a swing at the playground at which point is her kinetic energy increasing and her potential energy decreasing
Nimfa-mama [501]
Whenever an object is falling, its potential energy
is decreasing and its kinetic energy is increasing.

Olivia's potential energy is decreasing and her kinetic energy
is increasing as she moves toward the right side of the picture,
all the way from W, through X, to the bottom of the arc.
7 0
3 years ago
Read 2 more answers
What makes a model not useful?
lina2011 [118]
Answer is b hope this helps
8 0
3 years ago
A sinusoidally oscillating current I ( t ) with an amplitude of 9.55 A and a frequency of 359 cycles per second is carried by a
UNO [17]

Answer:

P_{avg} = 6.283*10^{-9} \ W

Explanation:

Given that;

I₀ = 9.55 A

f = 359 cycles/s

b = 72.2 cm

c = 32.5 cm

a = 80.2 cm

Using the formula;

\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})

where;

E= \frac{d \phi}{dt}

E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t

E_{rms} =   \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}

Replacing our values into above equation; we have:

E_{rms} =   \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}

E_{rms} =   \frac {8.98909588*10^{-4} }{\sqrt{2}}

E_{rms} =   6.356*10^{-4} \ V

Then the P_{avg is calculated as:

P_{avg} = \frac{E^2}{R}

P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}

P_{avg} = 6.283*10^{-9} \ W

6 0
3 years ago
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